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Arturiano [62]
2 years ago
6

What is 80% of 230? ​

Mathematics
2 answers:
Orlov [11]2 years ago
6 0

Answer:

184

Step-by-step explanation:

Stolb23 [73]2 years ago
5 0

0.80 x 230 = 184

You are welcome

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What is the number in standard form?<br><br> 4.9×10^8
belka [17]

In order to get the answer to this question you have to remember that if the exponent is positive you need to move to the right, and if the exponent is negative you got to move to the left.

  • 4.9\times10^8
  • Positive so move to the right 8 times....
  • 4.9\times10^8 = 490000000.
  • = 490000000.

Therefore the answer is "490000000."

Hope this helps!

Nonportrit

6 0
3 years ago
Factoring , <br> a² +4a - 5
givi [52]

the factors of the term a^2+4a-5 are (a-1)(a+5)

Step-by-step explanation:

We need to find factors of the term: a^2+4a-5

For finding factors:

We need to break the middle term such that their product is equal to product of first and last term and their sum is equal to middle term

Breaking the middle term and finding factors:

a^2+4a-5

=a^2+5a-a-5

=a(a+5)-1(a+5)

=(a-1)(a+5)

So, the factors of the term a^2+4a-5 are (a-1)(a+5)

Keywords: Finding factors

Learn more about finding factors at:

  • brainly.com/question/9045597
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  • brainly.com/question/13168205

#learnwithBrainly

6 0
2 years ago
The diameter of the sun is about is about 1 390 000 km;
sattari [20]

Answer:

1.39 × 10¹² or 1.39 × 10^12

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
What is the avarage rate from x=1 to x=3
guajiro [1.7K]

It depends on the line or parabola that you are talking about, more information would be needed to answer this correctly.

6 0
2 years ago
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How many different ways can you make 82 cents using current u.s. currency
andrew11 [14]
 <span>You can probably just work it out. 

You need non-negative integer solutions to p+5n+10d+25q = 82. 

If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80. 


So this is the same as n + 2d + 5q ≤ 16 

So now you simply have to "crank out" the cases. 

Case q=0 [ n + 2d ≤ 16 ] 

Case (q=0,d=0) → n = 0 through 16 [17 possibilities] 
Case (q=0,d=1) → n = 0 through 14 [15 possibilities] 
... 
Case (q=0,d=7) → n = 0 through 2 [3 possibilities] 
Case (q=0,d=8) → n = 0 [1 possibility] 

Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81 

Case q=1 [ n + 2d ≤ 11 ] 
Case (q=1,d=0) → n = 0 through 11 [12] 
Case (q=1,d=1) → n = 0 through 9 [10] 
... 
Case (q=1,d=5) → n = 0 through 1 [2] 

Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42 

Case q=2 [ n + 2 ≤ 6 ] 
Case (q=2,d=0) → n = 0 through 6 [7] 
Case (q=2,d=1) → n = 0 through 4 [5] 
Case (q=2,d=2) → n = 0 through 2 [3] 
Case (q=2,d=3) → n = 0 [1] 

Total from case q=2: 1 + 3 + 5 + 7 = 16 

Case q=3 [ n + 2d ≤ 1 ] 
Here d must be 0, so there is only the case: 
Case (q=3,d=0) → n = 0 through 1 [2] 

So the case q=3 only has 2. 

Grand total: 2 + 16 + 42 + 81 = 141 </span>
3 0
3 years ago
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